Failure,Analysis,and Design of Laminates 379 Using the following identity, Sin Cos kx≡ (5.14) 2 Sin 2 Then, cs20=0mW习 (5.15a) Cos40:=0frN≥3 (5.15b) Thus, An=Uh (5.16a) Similarly,it can be shown that A12=U4h, (5.16b) A22=11h, (5.16c) e(2出h (5.16d) Therefore, UU, 0 [A]= U 0 h (5.17) 0 u1-u4 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 379 Using the following identity,1 . (5.14) Then, (5.15a) (5.15b) Thus, . (5.16a) Similarly, it can be shown that , (5.16b) , (5.16c) . (5.16d) Therefore, (5.17) Cos Sin Sin kx N x x k N = ∑ ≡ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 1 1 2 2 2 1 2 Cos 2 0 1 1 θk k N for N = ∑ = ≥ Cos . 40 3 1 θk k N for N = ∑ = ≥ A Uh 11 1 = A Uh 12 4 = A Uh 22 1 = A U U 66 h 1 4 2 = ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ [] . A U U U U U U = h − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ 1 4 4 1 1 4 0 0 0 0 2 1343_book.fm Page 379 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
380 Mechanics of Composite Materials,Second Edition Because Equation(5.15b)is valid only for N>3,this proves that one needs at least three plies to make a quasi-isotropic laminate. For a symmetric quasi-isotropic laminate,the extensional compliance matrix is given by U U 0 u-u u-u? a U U 0 (5.18) u-u u-u 2 0 0 u1-u4] From the definitions of engineering constants given in Equations(4.35), (4.37),(4.39),(4.42),and (4.45),and using Equation (5.18),the elastic moduli of the laminate are independent of the angle of the lamina and are given by E.=E,=E=-1=U-Ui (5.19a) Aih U Gw=G=1=4-4 ΓA6h-21 (5.19%) Vxy=Vy=Viso=- (5.19c) 5.3 Failure Criterion for a Laminate A laminate will fail under increasing mechanical and thermal loads.The laminate failure,however,may not be catastrophic.It is possible that some layer fails first and that the composite continues to take more loads until all the plies fail.Failed plies may still contribute to the stiffness and strength of the laminate.The degradation of the stiffness and strength properties of each failed lamina depends on the philosophy followed by the user. When a ply fails,it may have cracks parallel to the fibers.This ply is still capable of taking load parallel to the fibers.Here,the cracked ply can be replaced by a hypothetical ply that has no transverse 2006 by Taylor Francis Group,LLC
380 Mechanics of Composite Materials, Second Edition Because Equation (5.15b) is valid only for N ≥ 3, this proves that one needs at least three plies to make a quasi-isotropic laminate. For a symmetric quasi-isotropic laminate, the extensional compliance matrix is given by . (5.18) From the definitions of engineering constants given in Equations (4.35), (4.37), (4.39), (4.42), and (4.45), and using Equation (5.18), the elastic moduli of the laminate are independent of the angle of the lamina and are given by (5.19a) (5.19b) (5.19c) 5.3 Failure Criterion for a Laminate A laminate will fail under increasing mechanical and thermal loads. The laminate failure, however, may not be catastrophic. It is possible that some layer fails first and that the composite continues to take more loads until all the plies fail. Failed plies may still contribute to the stiffness and strength of the laminate. The degradation of the stiffness and strength properties of each failed lamina depends on the philosophy followed by the user. • When a ply fails, it may have cracks parallel to the fibers. This ply is still capable of taking load parallel to the fibers. Here, the cracked ply can be replaced by a hypothetical ply that has no transverse [ ]* A h U U U U U U U U U U U = − − − − − − 1 0 1 1 2 4 2 4 1 2 4 2 4 1 2 4 2 1 1 2 U U U 4 2 1 4 0 0 0 2 − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ EEE A h U U U x y iso == = = 1 − 11 1 2 4 2 1 * , G G A h U U xy iso == = 1 − 2 66 1 4 * , ννν xy yx iso A A U U = = =− = 12 22 4 1 * * . 1343_book.fm Page 380 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 381 stiffness,transverse tensile strength,and shear strength.The longi- tudinal modulus and strength remain unchanged. When a ply fails,fully discount the ply and replace the ply of near zero stiffness and strength.Near zero values avoid singularities in stiffness and compliance matrices. The procedure for finding the successive loads between first ply failure and last ply failure given next follows the fully discounted method: 1.Given the mechanical loads,apply loads in the same ratio as the applied loads.However,apply the actual temperature change and moisture content. 2.Use laminate analysis to find the midplane strains and curvatures. 3.Find the local stresses and strains in each ply under the assumed load. 4.Use the ply-by-ply stresses and strains in ply failure theories dis- cussed in Section 2.8 to find the strength ratio.Multiplying the strength ratio to the applied load gives the load level of the failure of the first ply.This load is called the first ply failure load. 5.Degrade fully the stiffness of damaged ply or plies.Apply the actual load level of previous failure. 6.Go to step 2 to find the strength ratio in the undamaged plies: If the strength ratio is more than one,multiply the strength ratio to the applied load to give the load level of the next ply failure and go to step 2. If the strength ratio is less than one,degrade the stiffness and strength properties of all the damaged plies and go to step 5. 7.Repeat the preceding steps until all the plies in the laminate have failed.The load at which all the plies in the laminate have failed is called the last ply failure. The procedure for partial discounting of fibers is more complicated.The noninteractive maximum stress and maximum strain failure criteria are used to find the mode of failure.Based on the mode of failure,the appropriate elastic moduli and strengths are partially or fully discounted. Example 5.3 Find the ply-by-ply failure loads for a [0/90],graphite/epoxy laminate. Assume the thickness of each ply is 5 mm and use properties of unidirec- tional graphite/epoxy lamina from Table 2.1.The only load applied is a tensile normal load in the x-direction-that is,the direction parallel to the fibers in the 0°ply. 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 381 stiffness, transverse tensile strength, and shear strength. The longitudinal modulus and strength remain unchanged. • When a ply fails, fully discount the ply and replace the ply of near zero stiffness and strength. Near zero values avoid singularities in stiffness and compliance matrices. The procedure for finding the successive loads between first ply failure and last ply failure given next follows the fully discounted method: 1. Given the mechanical loads, apply loads in the same ratio as the applied loads. However, apply the actual temperature change and moisture content. 2. Use laminate analysis to find the midplane strains and curvatures. 3. Find the local stresses and strains in each ply under the assumed load. 4. Use the ply-by-ply stresses and strains in ply failure theories discussed in Section 2.8 to find the strength ratio. Multiplying the strength ratio to the applied load gives the load level of the failure of the first ply. This load is called the first ply failure load. 5. Degrade fully the stiffness of damaged ply or plies. Apply the actual load level of previous failure. 6. Go to step 2 to find the strength ratio in the undamaged plies: • If the strength ratio is more than one, multiply the strength ratio to the applied load to give the load level of the next ply failure and go to step 2. • If the strength ratio is less than one, degrade the stiffness and strength properties of all the damaged plies and go to step 5. 7. Repeat the preceding steps until all the plies in the laminate have failed. The load at which all the plies in the laminate have failed is called the last ply failure. The procedure for partial discounting of fibers is more complicated. The noninteractive maximum stress and maximum strain failure criteria are used to find the mode of failure. Based on the mode of failure, the appropriate elastic moduli and strengths are partially or fully discounted. Example 5.3 Find the ply-by-ply failure loads for a graphite/epoxy laminate. Assume the thickness of each ply is 5 mm and use properties of unidirectional graphite/epoxy lamina from Table 2.1. The only load applied is a tensile normal load in the x-direction — that is, the direction parallel to the fibers in the 0° ply. [0 90 / ]s 1343_book.fm Page 381 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
382 Mechanics of Composite Materials,Second Edition Solution Because the laminate is symmetric and the load applied is a normal load, only the extensional stiffness matrix is required.From Example 4.4,the extensional compliance matrix is 5.353×10-10 -2.297×10-1 0 [A]= -2.297×10119.886×10-10 0 Pa-m 0 0 9.298×10-9 which,from Equation(5.1a),gives the midplane strains for symmetric lam- inates subjected to N.=1 N/m as 5.353×10-10] -2.297×10-11 0 The midplane curvatures are zero because the laminate is symmetric and no bending and no twisting loads are applied. The global strains in the top 0o ply at the top surface can be found as follows using Equation(4.16), Ex 5.353×10-10 T01 -2.297×10-11 +(0.0075) 0 5.353×10-10 -2.297×10-1 0 Using Equation(2.103),one can find the global stresses at the top surface of the top 0°ply as 6￥ 181.8 2.897 5.353×10-10 6 2.897 10.35 0 (10) -2.297×10-1 0 0 7.17 0 2006 by Taylor Francis Group,LLC
382 Mechanics of Composite Materials, Second Edition Solution Because the laminate is symmetric and the load applied is a normal load, only the extensional stiffness matrix is required. From Example 4.4, the extensional compliance matrix is which, from Equation (5.1a), gives the midplane strains for symmetric laminates subjected to Nx = 1 N/m as . The midplane curvatures are zero because the laminate is symmetric and no bending and no twisting loads are applied. The global strains in the top 0° ply at the top surface can be found as follows using Equation (4.16), Using Equation (2.103), one can find the global stresses at the top surface of the top 0° ply as [ ] . . . . * A = × −× − × − − − 5 353 10 2 297 10 0 2 297 10 9 10 11 11 886 10 0 0 0 9 298 10 10 1 9 × × ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ − − . , Pa m- ε ε γ x y xy 0 0 0 10 5 353 10 2 297 10 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = × − × − . . − ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 11 0 ε ε γ x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = × − × − − 5 353 10 2 297 10 10 11 . . 0 0 0075 0 0 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ + ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ (. ) = × − × ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ − − 5 353 10 2 297 10 0 10 11 . . . σ σ τ x y xy top ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = 0° 181 8 2 897 0 2 897 1 , . . . 0 35 0 0 0 7 17 10 5 353 10 2 29 9 10 . . ( ) . . ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ × − − 7 10 0 11 × ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ − 1343_book.fm Page 382 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 383 9.726×101 1.313 Pa 0 Using the transformation Equation (2.94),the local stresses at the top surface of the top 0o ply are 1 0 9.726×10 02 0 1 0 1.313×10 00 0 9.726×10 1.313 Pa 0 All the local stresses and strains in the laminate are summarized in Table 5.1 and Table 5.2. TABLE 5.1 Local Stresses (Pa)in Example 5.3 Ply no. Position c 02 T12 1(0) Top 9.726×101 1.313×10° 0.0 Middle 9.726×101 1.313×10° 0.0 Bottom 9.726×101 1.313×100 0.0 2(90) Top -2.626×10° 5.472×10 0.0 Middle -2.626×10 5.472×100 0.0 Bottom -2.626×10° 5.472×10° 0.0 3(0) Top 9.726×101 1313×10° 0.0 Middle 9.726×101 1313×10° 0.0 Bottom 9.726×101 1.313×10° 0.0 TABLE 5.2 Local Strains in Example 5.3 Ply no. Position E n tn 1(0) Top 5.353×10-10 -2.297×10-11 0.0 Middle 5.353×1010 -2.297×10-11 0.0 Bottom 5.353×10-10 -2.297×10-11 0.0 2(90) Top -2.297×10-11 5.353×10-10 0.0 Middle -2.297×10-1 5.353×10-10 0.0 Bottom -2.297×10-11 5.353×10-10 0.0 3(0) Top 5.353×1010 -2.297×10-11 0.0 Middle 5.353×1010 -2.297×10-11 0.0 Bottom 5.353×10-10 -2.297×10-1 0.0 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 383 . Using the transformation Equation (2.94), the local stresses at the top surface of the top 0° ply are . All the local stresses and strains in the laminate are summarized in Table 5.1 and Table 5.2. TABLE 5.1 Local Stresses (Pa) in Example 5.3 Ply no. Position σ1 σ2 τ12 1 (0°) Top Middle Bottom 9.726 × 101 9.726 × 101 9.726 × 101 1.313 × 100 1.313 × 100 1.313 × 100 0.0 0.0 0.0 2 (90°) Top Middle Bottom –2.626 × 100 –2.626 × 100 –2.626 × 100 5.472 × 100 5.472 × 100 5.472 × 100 0.0 0.0 0.0 3 (0°) Top Middle Bottom 9.726 × 101 9.726 × 101 9.726 × 101 1.313 × 100 1.313 × 100 1.313 × 100 0.0 0.0 0.0 TABLE 5.2 Local Strains in Example 5.3 Ply no. Position ε1 ε2 τ12 1 (0°) Top Middle Bottom 5.353 × 10–10 5.353 × 10–10 5.353 × 10–10 –2.297 × 10–11 –2.297 × 10–11 –2.297 × 10–11 0.0 0.0 0.0 2 (90°) Top Middle Bottom –2.297 × 10–11 –2.297 × 10–11 –2.297 × 10–11 5.353 × 10–10 5.353 × 10–10 5.353 × 10–10 0.0 0.0 0.0 3 (0°) Top Middle Bottom 5.353 × 10–10 5.353 × 10–10 5.353 × 10–10 –2.297 × 10–11 –2.297 × 10–11 –2.297 × 10–11 0.0 0.0 0.0 = ⎡ × ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 9 726 10 1 313 0 1 . . Pa σ σ τ 1 2 12 0 100 010 001 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ °, top ⎥ × × ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = × 9 726 10 1 313 10 0 9 726 10 1 1 0 1 . . . .313 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ Pa 1343_book.fm Page 383 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC